The following equation is true for all real values of $n$ for which the expression on the left is defined, and $A$ is a polynomial expression. $\dfrac{n^2+24n+144}{n^2+11n-12}\cdot\dfrac{n^2+2n-3}{A}=1$ What is $A$ ? $A=$
Answer: The left side of the equation is a product of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting product on the left side should cancel out completely. In order to solve for $A$, let's multiply the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $n^2+24n+144$, of the first expression can be factored as $(n+12)(n+12)$ using the perfect square pattern. The denominator, $n^2+11n-12$, of the first expression can be factored as $(n+12)(n-1)$ using the sum-product pattern. The numerator, $n^2+2n-3$, of the second expression can be factored as $(n+3)(n-1)$ using the sum-product pattern. Now the product looks as follows: $\dfrac{(n+12)(n+12)}{(n+12)(n-1)}\cdot\dfrac{(n+3)(n-1)}{A}$ To find the product of two rational expressions, we multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=}\dfrac{(n+12)(n+12)}{(n+12)(n-1)}\cdot\dfrac{(n+3)(n-1)}{A} \\\\\\ &= \dfrac{(n+12)(n+12) \cdot (n+3)(n-1)}{(n+12)(n-1) \cdot A} &\text{Multiply across.}\\\\\\ &= \dfrac{(n+12){\cancel{(n+12)}} (n+3){\cancel{(n-1)}}}{{\cancel{(n+12)}}{\cancel{(n-1)}} A} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(n+12)(n+3)}{A} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{(n+12)(n+3)}{A}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $A=(n+12)(n+3)$, which is equivalent to $n^2+15n+36$.